- All are divible by 3,9. - They are exactly 99 less than the next rounded,I dont know what we call it. - Add 9 to each and they become divisible by 10 But is there a way to get this series?
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user23738Jan 26 '12 at 19:30

5 Answers
5

There is actually a straightforward reason. As $99$ is $1$ less than $100$, we get a fairly simple expression for its decimal $$\frac{1}{99}=0.01010101010101\overline{01}\dots$$ Now, $$\frac{1}{9801}=\left(\frac{1}{99}\right)^2,$$ and the decimal expansion follows from the formula for general power series $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$

Letting $x=\frac{1}{100}$, the decimal expansion for $\frac{1}{99}$ given above is exactly the same thing as writing $\frac{1}{99}=\sum_{n=1}^\infty x^n$. Applying our identity, the $x$ in front accounts for the double zero. Once $n$ is around $99$ we expect to miss a number because we are forcing things to be in decimal, and there will be carrying, which is why the number 98 is missed.

A similar pattern will occur for $\frac{1}{998001}=\left(\frac{1}{999}\right)^2,$ since as before $$\frac{1}{999}=0.001001\overline{001}.$$

$1/9801$ and the other numbers are all in the form of $$\frac{1}{999...9^2}.$$ You can see similar results for things like $1/9.9$ squared (only you get fewer zeros after the decimal). You can also try things like $1/9.999$ squared ($1/(9.999^2)$) and get more zeros between the magic numbers.

Now try $1/9800$ or $1/998000$ and see the magic number sequences that you get. :) I'm still not understanding that one.

Since this was recently brought back to the front page, I wanted to point out a particular numberphile blog video on this exact question.

In this video, Brady Haran interviews Dr. James Grime and they discuss why this happens and how to generalize the result.

If I were to briefly summarize the key idea: in the video, they show that this boils down to considering $\dfrac{12345679}{999999999} = \dfrac{1}{81}$, or to evaluating $\dfrac{1}{(1-x)^2}$ at $x = \frac{1}{10}$ and considering a sort of generating functions.

I was about to say:
Let's not forget that 9801 is a fundamental solution for x in $x^2-29y^2=1$

When I saw that there is an error in Wolframalpha's treatment of this equation. While 9801 which the program offers, is indeed a solution for x, is it not the fundamental solution (which is x=70 and y=13). This solution combined with itself Brahmagupta style gives x=9801 and y=1920